Two closed - end pipes , when sounded together produce ` 5 beats//s`. If their lengths are in the ratio `100 : 101` , then fundamental notes ( in Hz) produced by them are

Options

A. `245 , 250`
B. `250 , 255`
C. `495 ,500`
D. `500 , 505`.

Explanation

To solve the problem step by step, we need to find the fundamental frequencies of two closed-end pipes that produce 5 beats per second when sounded together. The lengths of the pipes are in the ratio of 100:101.

### Step 1: Understand the relationship between frequency and length
The fundamental frequency (F) of a closed-end pipe is inversely proportional to its length (L). This can be expressed as:
\[ F \propto \frac{1}{L} \]
Thus, for two pipes with lengths \( L_1 \) and \( L_2 \), the frequencies \( F_1 \) and \( F_2 \) can be written as:
\[ \frac{F_1}{F_2} = \frac{L_2}{L_1} \]

### Step 2: Set up the lengths based on the given ratio
Let the lengths of the two pipes be:
- \( L_1 = 100 \) (for the first pipe)
- \( L_2 = 101 \) (for the second pipe)

### Step 3: Express the frequencies in terms of one variable
Using the inverse relationship, we can express the frequencies as:
\[ F_1 = k \cdot \frac{1}{L_1} = \frac{k}{100} \]
\[ F_2 = k \cdot \frac{1}{L_2} = \frac{k}{101} \]
where \( k \) is a constant.

### Step 4: Use the beat frequency information
The beat frequency is given as 5 beats per second, which means:
\[ |F_1 - F_2| = 5 \]

### Step 5: Substitute the expressions for frequencies
Substituting the expressions for \( F_1 \) and \( F_2 \):
\[ \left| \frac{k}{100} - \frac{k}{101} \right| = 5 \]

### Step 6: Simplify the equation
To simplify, we can find a common denominator:
\[ \left| \frac{101k - 100k}{100 \cdot 101} \right| = 5 \]
This simplifies to:
\[ \left| \frac{k}{100 \cdot 101} \right| = 5 \]

### Step 7: Solve for \( k \)
Multiplying both sides by \( 100 \cdot 101 \):
\[ k = 5 \cdot 100 \cdot 101 = 50500 \]

### Step 8: Calculate the fundamental frequencies
Now we can find \( F_1 \) and \( F_2 \):
\[ F_1 = \frac{50500}{100} = 505 \, \text{Hz} \]
\[ F_2 = \frac{50500}{101} = 500 \, \text{Hz} \]

### Step 9: Final answer
The fundamental frequencies produced by the two pipes are:
- \( F_1 = 505 \, \text{Hz} \)
- \( F_2 = 500 \, \text{Hz} \)